Algebra Skillbuilder Solutions:
1. Starting with
, you’ll need to find a common denominator to add/subtract the
fractions. If you choose the common denominator 15, you can multiply each fraction by one
accordingly to achieve that denominator:
(
yields
.
Combining the terms in the numerator you arrive at the answer:
2. Beginning with
alone:
, factor out the 4s on the left hand side of the equation to get
Then divide both sides by 4 and you’ll have the answer:
3. Starting with
remember Order of Operations and deal with the parentheses
first. Multply the exponent 2 across the parentheses to yield:
Then you can combine the m terms, adding the exponents because you’re multiplying their
bases:
4. Beginning with:
, your job is to combine like terms to solve for a.
It’s likely easiest to begin by combining on each side of the equals sign:
5a – 5a = 3a
0 = 3a
So a must equal 0.
5. As you start with
, your immediate goal should be to get rid of denominators to
have a straight linear equation to work with. In order to do that, multiply both sides of the
equation by 5:
Then distribute that multiplication across the parentheses:
4c + 10 = 5c – 10
Then subtract 4c from both sides:
10 = c – 10
Then add 10 to both sides:
20 = c
6. Here it’s important to keep your goal in mind, which isn’t to solve for any one variable but
rather for the combination . You start with two equations:
So your goal should first be to get rid of the denominator in the first equation by multiplying
both sides by y:
x + y = 3y
Then, as your goal is to solve for a combination of x and z, you’ll want to y in terms of x so as to
eliminate the y variable. Subtract y from both sides:
x = 2y
Then divide both sides by 2 and you’ll have solved for y:
This means that in the other equation, you can plug in instead of y:
Then subtract from each side to get the x term alone:
z = ½ (x)
And then divide both sides by x to get the fraction to match the question, and you’ll find that
.
7. You begin this problem with two equations:
And your goal is to solve for b. Your first goal should be to reduce the number of variables, and
the second equation affords you a quick opportunity to do so. Since it tells you that b = 2a, you
can plug 2a in for b in the first equation:
4a – 2a = 6
Meaning that 2a = 6 and a = 3. Since b = 2a, then b = 2(3), so b = 6.
8. This problem tests your familiarity with exponent rules, a very important concept on the GMAT.
As you start with
as
, you should recognize that a negative exponent like
is the same thing
- essentially you can take the negative exponent, flip it over the fraction bar, and make it
positive. So what you really have here is
. And with that in mind, you know that if
you’re multiplying together the same base (in this case 5) to two different exponents, you add
the exponents, giving you
.
9. The answer here is 0. Remember, = 1, so even though 7 is not a factor of 31, because 1 is a
factor of 31 you can still make satisfy the equation. Accordingly, the correct answer is 0.
10. Beginning with
, look for squares within the numbers underneath the radical and you’ll see
that what you really have is:
, which means that the answer is
.
11. With root problems that involve addition or subtraction, you’ll typically want to factor out the
roots to eliminate squares within them. As you have
. This allows you to take the square root of the 32 terms,
actual squares:
leaving:
, look for ways to pull out
. Combine the like
terms and you’ll find the answer: 3 + 4
.
12. With this problem you should recognize that you can quickly calculate the square root portion,
as you know that
. That means that
is really just
, which you know is 8.
13. Remember that with inequalities, you can add and subtract values from either side the same
exact way you can with equations (the only time things get tricky is when you multiply or divide
by a negative, or a variable). So as you start with:
, you can subtract 3a from each side to get:
, then to get a alone subtract 6 from both sides, and you’ll see that:
14.
means that either:
, or
Solving for each, if 2x + 3 = 5, then subtract 3 from each side to get:
2x = 2, and then divide both sides by 2 to get:
x=1
Or, if 2x + 3 = -5, then subtract three from each side to get:
2x = -8, then divide both sides by 2 to get:
x = -4
Looking at the answer choices, you’ll only find -4, so -4 is correct.
15. In a quadratic like
, you want to factor out the numeric term (in this case 12,
but don’t forget that it’s negative). 12 could be:
1 and 12
2 and 6
3 and 4
Then look for a combination that adds/subtracts to the middle number (4a). Since you need one
factor to be negative and one to be positive, the combination that works to add to 4 is (a + 6)(a
– 2) = 0. But remember – you still have to solve the equation. If a = -6 or if a = 2, the equation
stands, so the solutions are 2 and -6.
16. You can split
clean. You know that:
into two inequalities to help keep your algebraic manipulation
3y + 3 > -6, which allows you to subtract 3 from both sides to get:
3y > -9, and then you can divide both sides by 3 to get:
y > -3
So that’s one part of the solution, and now you can attack the other half. Knowing that:
3y + 3 < 12 allows you to again subtract 3 from both sides to get:
3y < 9, whereby you can again divide by 3 to get:
y<3
Combining what you know, you know that -3 < y < 3.
17. When you’re given two inequalities like x + y > 8 and x – y < 4, if you can get the signs going in
the same direction you can add the inequalities together to eliminate a variable, essentially
using the “Elimination Method” for multiple equations and multiple variables. And since the
question asks you about y, your goal should be to eliminate x. You can do that by taking the
second inequality:
x – y < 4, and multiplying everything by -1 to flip the sign:
-x + y > -4. Pair that with the other inequality:
x + y > 8, and add them together
2y > 4
So y > 2
18. You should recognize the opportunity to use the Difference of Squares rule here, as the given
equation fits the form x2 – y2 = 0:
4x2 – 9 = 0 can be written as (2x)2 – 32= 0 This allows you to transform this into:
(2x + 3)(2x – 3) = 0, matching answer choice B.
19. This is a great question for using the answer choices as guides, since the question asks for what
“could be” the value of x + y and not what “is” the value of x + y, the algebra won’t necessarily
lead to a specific value. You’re looking for a number that equals its own square root, and since 1 and 2 don’t qualify, the answer must be 0.
20. Solving for c given the equation
adding 27 to both sides to form:
, you’ll first want to create a quadratic by
. Then you can divide all terms by 3:
, and now you can either factor or recognize the common algebraic equation
for squares:
, so c must be -3.